Optimal. Leaf size=231 \[ -\frac{63 i}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{63 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^2 c^{5/2} f}-\frac{21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.235862, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ -\frac{63 i}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{63 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^2 c^{5/2} f}-\frac{21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3522
Rule 3487
Rule 51
Rule 63
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{\int \frac{\cos ^4(e+f x)}{\sqrt{c-i c \tan (e+f x)}} \, dx}{a^2 c^2}\\ &=\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{\left (9 i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{8 a^2 f}\\ &=\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(63 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^2 f}\\ &=-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(63 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 f}\\ &=-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac{(63 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^2 c f}\\ &=-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{63 i}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(63 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^2 c^2 f}\\ &=-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{63 i}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(63 i) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{128 a^2 c^2 f}\\ &=\frac{63 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^2 c^{5/2} f}-\frac{63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{63 i}{128 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 5.78185, size = 182, normalized size = 0.79 \[ \frac{\sec ^2(e+f x) \sqrt{c-i c \tan (e+f x)} (\sin (3 (e+f x))-i \cos (3 (e+f x))) \left (-141 i \sin (e+f x)-159 i \sin (3 (e+f x))-18 i \sin (5 (e+f x))-547 \cos (e+f x)+31 \cos (3 (e+f x))+2 \cos (5 (e+f x))+315 e^{-i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )\right )}{1280 a^2 c^3 f (\tan (e+f x)-i)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.041, size = 159, normalized size = 0.7 \begin{align*}{\frac{-2\,i{c}^{3}}{f{a}^{2}} \left ({\frac{1}{16\,{c}^{5}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{15}{16} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{17\,c}{8}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }-{\frac{63\,\sqrt{2}}{32}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) }+{\frac{3}{16\,{c}^{5}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}+{\frac{1}{16\,{c}^{4}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{40\,{c}^{3}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.53278, size = 999, normalized size = 4.32 \begin{align*} \frac{{\left (315 i \, \sqrt{\frac{1}{2}} a^{2} c^{3} f \sqrt{\frac{1}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (4032 i \, a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 4032 i \, a^{2} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{4} c^{5} f^{2}}} + 4032 i\right )} e^{\left (-i \, f x - i \, e\right )}}{4096 \, a^{2} c^{2} f}\right ) - 315 i \, \sqrt{\frac{1}{2}} a^{2} c^{3} f \sqrt{\frac{1}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-4032 i \, a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 4032 i \, a^{2} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{4} c^{5} f^{2}}} + 4032 i\right )} e^{\left (-i \, f x - i \, e\right )}}{4096 \, a^{2} c^{2} f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-8 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 64 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 344 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 203 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 95 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 10 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{1280 \, a^{2} c^{3} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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